JEE Main Study Notes for Coordinate Geometry: Important Formulas, Study Tips & Previous Year Solved Questions

    Urvashi Singhal Urvashi Singhal
    Exams Prep Master

    Coordinate Geometry is one of the most interesting and easiest topics in JEE Main syllabus. It carries maximum weightage (from 17-20%) in Mathematics section, thereby raising the scope for test takers to secure a high rank in JEE Main.

    Read JEE Main Syllabus for Mathematics

    • Considering last 5 years JEE Main question papers, around 7-8 questions have come from coordinate geometry out of 30 questions (as per previous pattern).
    • Around 50% of the questions of the section are from straight lines and circles.
    • Consistently 6-8 questions have appeared from the section in JEE Advanced too.

    Since it is a vast topic, it can be further divided into parts like parabola, ellipse, and hyperbola. The section consists of many formulas, which if memorized can help in instant solving of problems. 

    If a candidate wishes to ace JEE Main, the catch to master coordinate geometry is to look beyond CBSE syllabus. A thorough study of NCERT is widely recommended by all JEE test takers. Read the article for reference to important formulas in Coordinate Geometry and reference material.

    Important topics in Coordinate Geometry for IIT JEE

    2D GeometryDistance formula
    Section formula
    Area of triangle
    Locus of a point
    Transformation of axis
    Straight LinesConcepts of straight line
    Conic SectionCircle
    Pair of straight lines

    Locus of a Point

    Locus of a Point

    When a point moves so as to satisfy a given condition, or conditions, the path it traces out is called its Locus under these conditions. Locus of a point is a point, curve or region. 

    1. Representation of a point in two-dimensional space by coordinates:

    i. Coordinates of a point P in two-dimensional space w.r.t. OXY axes is an ordered pair of real numbers written as (x, y) such that the coordinates are the distances from the origin of the feet of the perpendiculars from the point P on the respective coordinate axes.

    ii. Coordinates of origin is (0,0).

    2. Equation of a curve/ region:

    i. The equation of a curve/ region is the relation which exists between the coordinates of every point on the curve/ region, and which holds for no other points except those lying on the curve/ region.

    ii. Equation of x-axis is y=0 ; equation of y-axis is x=0. 

    Question For Practice- Plot locus on x - y plane: 2y ≥ x2, y ≤ -2x2 + 3x

    Distance Formula

    Distance Formula

    Case I: Distance ‘d’ between any two points A ( x1, y1) and B ( x2, y2) on the coordinate axis is given by:

    Case II: Distance of a point ( x1, y1) from origin is given by:

    Question For Practice- A line is of length 10 and one end is at the point (2, -3); if the abscissa of the other end be 10, prove that its ordinate must be 3 or -9.

    Section Formula

    Section Formula

    Case I: Coordinates of the point which divides the line segment joining the points ( x1, y1) and ( x2, y2) in a given ratio m1:m2 internally is:

    Case II: Coordinates of the point which divides the line segment joining the points ( x1, y1) and ( x2, y2) in a given ratio m1:m2 (m1≠m2) externally is:

    Question For Practice- Find the ratio in which the line segment joining the points 3,4 and 1,2 is divided by y axis. {Ans. 3:1}

    Area of Triangle

    Area of Triangle

    Case I: Area of a triangle whose vertices are ( x1, y1), ( x2, y2), and (x3, y3) is:

    Question For Practice- Find the area of the quadrilateral the coordinates of whose vertices, taken in order, are (-1, 6), (-3, -9), (5, -8) and (3, 9). {Ans. 96}

    Equations of a Straight Line

    Equations of a Straight Line

    Case I: Point – Slope form

    Let ‘m’ be the slope of line and line is passing through point A( x1, y1). Then equation of line is

    Case II: Two point form

    where ( x1, y1) and ( x2, y2) are two given points on the line and m = y2 - y1 ∕ x2 - x1

    Case III: Intercept form

    Let the line make an intercept ‘a’ and ‘b’ on x-axis and y-axis respectively. Then equation of line is

    Case IV: Normal form

    Let the distance of line from the origin is ‘p’ and the angle it made with the origin is .Then equation of line is

    Question For Practice-

    Q1. Find the equation to the straight line which passes through the point (5, 6) and has intercepts on the axes

    i. equal in magnitude and both positive

    ii. equal in magnitude but opposite in sign

    Q2. If the point 5,2 bisects the intercept of a line between the axes, then find the equation of the line.

    Conic Section

    Conic Section

    Recognition of Conics

    General form: ax2+2hxy+by2+2gx+2fy+c = 0

    0> 0Real and distinct lines
    = 0Parallel lines
    < 0Imaginary lines
    ≠ 0= 0Parabola
    < 0Ellipse
    > 0Hyperbola
    > 0 and a + b = 0Rectangular Hyperbola
    < 0 and h = 0, a = bCircle


    Let P be any moving point and S be the focus (fixed point) of the conic. Let PM be the perpendicular distance of the point from the directrix (fixed line) of the conic. Then eccentricity or contracity ‘e’ of the conic is defined by

    Note: ‘e’ can never be negative.

    Ellipse0 < e < 1
    Parabolae = 1
    Hyperbolae > 1
    Circlee → 0
    Pair of straight linee → ∞



    A circle is the locus of a point which moves in a plane such that its distance from a fixed point in that plane is always constant. The fixed point is called the center of the circle and the constant distance is called the radius of the circle. 

    General Equationx2 + y2 + 2gx + 2fy + c = 0
    Center(-g, -f)
    Radiusg2 + f2 - c
    X - intercept2g2 - c
    Y - intercept2f2 - c
    Tangent (Point Form)xx1 + yy1 = a2
    Tangent (Slope Form)y = mx ± a1 + m2
    Tangent (Parametric Form)xcosθ + ysinθ = a

    Question For Practice- Find the equation to the circle which passes through the points (1, 2) and (4, 3) and which has its centre on the straight line 3x + 4y = 7. {Ans. 15x2 + 15y2 - 94x + 18y + 55 = 0}



    A conic section is the locus of a point such that its distance from a fixed point (S) always bears a constant ratio (e) to its distance from a fixed line (D).

    Properties of standard Parabola

    Equationy2 = 4axy2 = -4axx2 = 4ayx2 = -4ay
    Focus(a, 0)(-a, 0)(0, a)(0, -a)
    Equation of Directrixx + a = 0x - a = 0y + a = 0y - a = 0
    Extrimite of Latus Rectum(a, ±2a)(-a, ±2a)(±2a, a)(±2a, -a)
    Length of Latus Rectum4a4a4a4a
    Parametric Coordinates(at2, 2at)(-at2, 2at)(2at, at2)(2at, -at2)
    Focal Distance of P (x1, y1)Ia + x1IIa - x1IIa + y1IIa - y1I
    Equation of Tangent (Slope Form)y = mx + a/my = mx - a/my = mx - amy = mx + am2
    Equation of Tangent (Parametric Form)y = 1tx + aty = 1tx - aty = 1tx - a1t2y = 1tx + a1t2
    Equation of Normaltx + y = 2at + at3tx + y = 2at - at3tx + y = 3at2tx + y = at2



    An ellipse has two vertices, denoted by A and A’. The mid-point of the two vertices is called the center of the ellipse, denoted by C. The line segment AA’ is called the major axis of the ellipse.

    Properties of standard Ellipses

    Equationx2a2+ y2b2= 1x2a2+ y2b2= 1x2b2+ y2a2= 1
    Axisy = 0x = 0x = 0
    Major Axis (M)2a2b2a
    Minor Axis (m)2b2a2b
    Eccentricity1- b2/a21- a2/b21- b2/a2
    Foci(±ae, 0)(0, ±be)(0, ±ae)
    Directrixx = ±a/ey = ±b/ey = ±a/e
    Centre(0, 0)(0, 0)(0, 0)
    Latus Rectum Extremities(±ae, ±b2/a)(a2/b, ±be)(b2/a, ±ae)
    Latus Rectum2b2/a2a2/b2b2/a
    Focal Distancea - ex1, a + ex1b - ey1, b + ey1a - ey1, a + ey1
    Area bounded by Ellipseπabπabπab
    Parametric Coordinates(acosα, bsinα)(bcosα, asinα)(acosα, bsinα)



    A hyperbola has two vertices, denoted by A and A’. The mid-point of the two vertices is called the center of the hyperbola, denoted by C. The line segment AA’ is called the transverse axis of the hyperbola.

    Properties of standard Hyperbola

    Equationx2a2- y2b2= 1y2a2- x2b2= 1y2b2- x2a2= 1
    Vertices(±a, 0)(0, ±a)(0, ±b)
    Centre(0, 0)(0, 0)(0, 0)
    Transverse Axis2a2a2b
    Conjugate Axis2b2b2a
    Eccentricity (e)1+ b2/a21+ a2/b21+ a2/b2
    Foci(±ae, 0)(0, ±ae)(0, ±be)
    Directrixx = ±a/ex = ±a/ex = ±b/e
    Parametric Coordinates(asecθ, btanθ)(btanθ, asecθ)(atanθ, bsecθ)

    Tips to study Coordinate Geometry

    Tips to study Coordinate Geometry

    The chances of committing mistakes in coordinate geometry students are less. However certain students tend to make silly mistakes due to lack of conceptual clarity and less input time to the topics. Coordinate Geometry is an extremely scoring topic in JEE mathematics which can get a candidate a better rank in the JEE Main. Below mentioned are some tips which a candidate must follow to score high in this section:

    • Coordinate geometry problems should not be practiced without drawing figures.
    • A different notebook should be prepared for the formulas. Keep jotting down all important formulas that you come across while studying.
    • Memorize as many formulas as you can.
    • A careful reading of the question is required to understand what it is demanding, which will in a way save time.
    • Avoid referring to different books for the same concept as it will lead to confusion. Stick to one standard book and one reference book.

    Click here to read preparation tips for JEE Main Mathematics

    Previous Year Solved Questions

    Previous Year Solved Questions

    Question: A variable line passes through a fixed point P. The algebraic sum of the perpendicular drawn from (2, 0), (0, 2) and (1, 1) on the line is zero, then what are the coordinates of the P?

    Solution: Let P (x1, y1), then the equation of the line passing through P and whose gradient is m, is y − y1 = m (x − x1). Now according to the condition

    [{−2m + (mx1 − y1)} / {√1 + m2}]+ [{2 + (mx1 − y1)} / {√1 + m2} + {1 − m + (mx1 − y1)} / {√1 + m2}=0

    3 − 3m + 3mx− 3y1 = 0

    ⇒ y− 1 = m (x1 − 1)

    Since it is a variable line, so hold for every value of m.

    Therefore, y= 1, x= 1

    ⇒ P(1,1)

    Question: The area of a parallelogram formed by the lines ax ± by ± c = 0, is __________.

    Solution: ax ± by ± c = 0

    ⇒ x / [± c / a] + y/ [± c / b] = 1 which meets on axes at A (ca, 0), C (−ca, 0), B (0, cb), D (0, −cb).

    Therefore, the diagonals AC and BD of quadrilateral ABCD are perpendicular, hence it is a rhombus whose area is given by = [1 / 2] AC * BD = [ 1 / 2] * [2c / a] * [2c / b] = 2c2 / ab.

    Question: The equation of the lines, which passes through the point (3, – 2) and are inclined at 60o to the line √3x + y = 1 ____________.

    Solution: The equation of any straight line passing through (3, 2) is y + 2 = m (x − 3) …….(i) The slope of the given line is −√3.

    So, tan 60= ± [m − (−√3)] / [1 + m (−√3)]

    On solving, we get m = 0 or √3

    Putting the values of m in (i), the required equation of lines are

    y + 2 = 0 and √3x − y = 2 + 3√3.

    Question: The line 2x + 3y = 12 meets the x-axis at A and y-axis at B. The line through (5, 5) perpendicular to AB meets the x-axis, y-axis and the AB at C, D and E, respectively. If O is the origin of coordinates, then the area of OCEB is _______.

    Solution: Here O is the point (0, 0).

    The line 2x + 3y = 12 meets the y-axis at B and so B is the point (0, 4).

    The equation of any line perpendicular to the line 2x + 3y = 12 and passes through (5, 5) is 3x − 2y = 5 ……(i)

    The line (i) meets the x-axis at C and so coordinates of C are (5 / 3, 0).

    Similarly, the coordinates of E are (3, 2) by solving the line AB and (i).

    Thus, O (0, 0), C (5 / 3, 0), E (3, 2) and B (0, 4).

    Now the area of figure OCEB = area of ΔOCE + area of ΔOEB = [23 / 3] sq.units.



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